Download slides

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript
6.033, Spring 2014
TCP Congestion Control
Dina Katabi & Sam Madden
nms.csail.mit.edu/~dina
Sharing the Internet
How do you manage resources in a huge system
like the Internet, where users with different
interests share the same links?
Difficult because of:

Size


Billions of users, links, routers
Heterogeneity

bandwidth: 9.6Kb/s (then modem, now cellular), 10 Tb/s

latency: 50us (LAN), 133ms (wired), 1s (satellite), 260s (Mars)
Congestion
S1
10Mb/s
R1
S2


Why a problem?


100Mb/s
Sources compete for link
bandwidth, and buffer space



S1
S2
Sources are unaware of current state of resource
Sources are unaware of each other
Manifestations:

2Mb/s
D
Lost packets (buffer overflow at routers)
Long delays (queuing in router buffers)
In many situations will result in < 2 Mb/s of throughput for the
above topology (congestion collapse)
Objectives of Congestion Control
Efficiency & Fairness


Efficiency

Maximize link utilization

Minimize queue size (i.e., delay)

Minimize packet drops
Many solutions!


(S1= 1 Mb/s, S2= 1 Mb/s) and (S1=1.5
Mb/s, S2=0.5 Mb/s) are both
efficient.
Want Fairness
S1
10Mb/s
R1
S2
100Mb/s
2Mb/s
D
Fairness
 Max-Min


Fairness
At each bottleneck,
user gets min(user’s
demand, fair share)
User’s rate is the
minimum max-min fair
rate along the path
S1
S2
S2
R1 9Mb/s D
Demands
1= 1Mb/s
2= 7Mb/s
3= 
Max-min
Fair Rates
1= 1Mb/s
2= 4Mb/s
3= 4Mb/s
TCP
TCP


TCP provides reliability & congestion control
Reliable transmission ensures the receiver’s
application receives the correct and complete
data sent by the sender’s application



TCP recovers from lost packets, eliminates
duplicates and ensures in-order packet delivery to
the application
Reliability was discussed in 6.02
Congestion control

Sender reacts to congestion and discovers its fair
and efficient send rate
TCP Cong. Cont.

Basic Idea:


Send a few packets. If a packet is dropped
decrease rate. If no drops, increase rate
How does TCP detect drops?


Packets have sequence numbers
Receiver acks the next expected sequence number (i.e., if
received 1, it acks 2 saying it is expecting 2 to arrive next.
Note that TCP implementations ack bytes but for simplicity we
talk about acking packets.)
TCP controls throughput via the
congestion window


Congestion window is the number of outstanding
packets, i.e., number of packets sender can send
without waiting for an ack
TCP is window-based: sources change their sending
rate by modifying the window size: “cwnd”


Avg. Throughput = (Avg. cwnd) /(Avg. RTT)
Why not changing rate directly?

Window protocols are easy to implement (no need for
accurate timers, i.e., works for slow machines an sensors)
How much should TCP Increase/decrease?


Probe for the correct sending window
Additive Increase / Multiplicative
Decrease (AIMD)

Every RTT:
No loss: cwnd = cwnd + 1
A loss:
cwnd = cwnd /2
Additive Increase
Src
cwnd += 1
cwnd = 2
cwnd = 1
D
A
D D
cwnd = 3
A A
D D
D A A
Dest
Actually,
On ack arrival: cwnd = cwnd + 1/cwnd
On timeout:
cwnd = cwnd /2
cwnd = 4
A
AIMD Leads to Efficiency and Fairness
fairness
line
User 2: x2
(x1,x2)
(bx1+a,bx2+a)
(bx1,bx2)
Efficiency line
(x1+x2 = BW)
User 1: x1
TCP AIMD
Congestion
Window
Timeout because
of a packet loss
correct
cwnd
Grab back
Bandwidth
Time
Halve Congestion
Window (and Rate)
Need the queue to absorb these saw-tooth oscillations
“Slow Start”
Cold start a connection at startup or after a timeout

At the beginning of a connection, increase
exponentially


Src
On ack arrival: cwnd += 1
1
D
2
A
D D
4
A A
D D
8
D
A
Dest
D
A
A
A
A
A
A
A
Adding Slow Start
Congestion
Window
Loss+Timeout
AIMD
Time
Exponential
Increase
Slow start stops at halve
previous cwnd
Tweaking TCP
Fast Retransmit



Timeouts are too slow
When packet is dropped, receiver still acks the
next in-order packet
Use 3 duplicate ACKs to indicate a drop

Why 3? When this does not work?
Fast Recovery



If there are still ACKs coming in then no need for
slow-start
Divide cwnd by 2 after fast retransmit
Increment cwnd by 1/cwnd for each dupack
Putting It Together
Congestion
Window
Time
Putting It Together
Congestion
Window
Time
Slow Start
Putting It Together
Congestion
Window
3 dupacks (Fast Retransmit)
Time
Putting It Together
Congestion
Window
Time
Fast Recovery
TCP Steady-State Throughput as
Function of Loss Rate
8
1  Wm 
 Wm 
1 drop every  
 
 pk t so, drop rate is: p 
3Wm2
2  2 
 2 
2
Throughput  is the
packets sent divided by
the time it took to send
them
2
W(t)
1 drop
Wm
3Wm

4 RTT
From the two eq.

3/ 2
RTT p
Wm/2
RTT * Wm/2
time
Reflections on TCP
The probing mechanism of TCP is based on
causing congestion then detecting it
 Assumes that all sources cooperate
 Assumes flows are long enough
 Too bursty
 Vulnerable to non-congestion related loss (e.g.
wireless errors)
 Unfair to long RTT flows
